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\begin{document}

\sloppy

\needspace{4\baselineskip}
\section*{
Exercice 1}

 Dans le triangle LCT rectangle en L, on sait que :
\begin{itemize}
\item LC=2,4 cm
\item \(\widehat{LCT}\) = 57\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [LT]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle LWC rectangle en L, on sait que :
\begin{itemize}
\item LW=1,6 cm
\item WC=8,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{LCW}\).

\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle CFZ rectangle en C, on sait que :
\begin{itemize}
\item FZ=1,9 cm
\item \(\widehat{CFZ}\) = 55\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [CF]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 4}

 Dans le triangle CAJ rectangle en C, on sait que :
\begin{itemize}
\item CA=1,8 cm
\item AJ=7,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{CAJ}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle CSN rectangle en C, on sait que :
\begin{itemize}
\item CS=0,8 cm
\item \(\widehat{CSN}\) = 68\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [NS]. (Arrondir au dixième)

\newpage
\section*{\textcolor{red}{Correction}}

\needspace{4\baselineskip}
\subsection*{
Exercice 1}

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\end{tikzpicture}\end{center}


 Dans le triangle LCT rectangle en L, on cherche une relation entre l'angle aigu   \(\widehat{LCT}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{LCT}\)) = \(\dfrac{LT}{LC}\)
\par\addvspace{.5em}
d'où tan(57\textdegree{}) =\(\dfrac{LT}{2{,}4}\)
\par\addvspace{.5em}
On a donc LT = 2,4 \(\times\) tan(57\textdegree{}) \(\approx\) 3,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle LWC rectangle en L, on cherche une relation entre l'angle aigu   \(\widehat{LCW}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{LCW}\)) = \(\dfrac{LW}{WC}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{LCW}\)) =\(\dfrac{1{,}6}{8{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{LCW}\) = ArcSin(\(\dfrac{1{,}6}{8{,}5}\)) \(\approx\) 11\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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 Dans le triangle CFZ rectangle en C, on cherche une relation entre l'angle aigu   \(\widehat{CFZ}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{CFZ}\)) = \(\dfrac{CF}{FZ}\)
\par\addvspace{.5em}
d'où cos(55\textdegree{}) = \(\dfrac{CF}{1{,}9}\)

On a donc CF = 1,9 \(\times\) cos(55\textdegree{}) \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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 Dans le triangle CAJ rectangle en C, on cherche une relation entre l'angle aigu   \(\widehat{CAJ}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{CAJ}\)) = \(\dfrac{CA}{AJ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{CAJ}\)) = \(\dfrac{1{,}8}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{CAJ}\) =ArcCos(\(\dfrac{1{,}8}{7{,}9}\)) \(\approx\)77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.423,2.371) {68°};
\end{tikzpicture}\end{center}


 Dans le triangle CSN rectangle en C, on cherche une relation entre l'angle aigu   \(\widehat{CSN}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{CSN}\)) = \(\dfrac{CS}{SN}\)
\par\addvspace{.5em}
 d'où cos(68\textdegree{}) = \(\dfrac{0{,}8}{SN}\)
\par\addvspace{.5em}
On a donc SN = \(\dfrac{0{,}8}{\cos(68^{\circ})}\)\(\approx\) 2,1 cm\end{document}
